Question

Let $f(x) = \int_0^x g(t) \log_e \left( \frac{1 - t}{1 + t} \right) dt$, where $g$ is a continuous odd function. If $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( f(x) + \frac{x^2 \cos x}{1 + e^x} \right) dx = \left( \frac{\pi}{\alpha} \right)^2 - \alpha,$ then $\alpha$ is equal to .....

Answer 1

Given the function:

$f(x) = \int_0^{\infty} g(t) \ln\left(1 - \frac{t}{1+t} \right) dt,$

where $g(t)$ is a continuous odd function. To determine whether $f(x)$ is odd, compute $f(-x)$:
$f(-x) = \int_0^{\infty} g(t) \ln\left(1 - \frac{-t}{1+t} \right) dt.$

Using a substitution $t = -y$, we get:

$f(-x) = \int_0^{\infty} g(-y) \ln\left(1 + \frac{y}{1-y} \right) (-dy).$

Since $g(y)$ is odd ($g(-y) = -g(y)$) and the logarithmic term changes sign due to the negative argument:

$f(-x) = -\int_0^{\infty} g(y) \ln\left(1 - \frac{y}{1+y} \right) dy = -f(x).$

This shows that $f(x)$ is also an odd function:

$f(-x) = -f(x).$

Now consider the integral:

$I = \int_{- \pi/2}^{\pi/2} f(x) \cdot \frac{x^2 \cos x}{1 + e^x} dx.$

Using the odd nature of $f(x)$ and the even nature of $\frac{x^2 \cos x}{1 + e^x}$, the product $f(x) \cdot \frac{x^2 \cos x}{1 + e^x}$ is odd. Hence, the integral over symmetric limits simplifies as:

$I = \int_{-\pi/2}^{\pi/2} (\text{odd function}) dx = 0.$

To simplify further, consider the integral:

$I = 2 \int_0^{\pi/2} f(x) \cdot \frac{x^2 \cos x}{1 + e^x} dx.$

Now substitute and evaluate:

$f(x) = \int_0^{\infty} g(t) \ln\left(1 - \frac{t}{1+t}\right) dt.$

Substituting into the main integral, and evaluating using standard properties of trigonometric integrals, we simplify $I$ as:

$I = -\int_0^{\pi/2} x^2 \cos x \, dx.$

Let:

$I = \int_0^{\pi/2} x^2 \cos x \, dx = \left(x^2 \sin x \right)_0^{\pi/2} - \int_0^{\pi/2} 2x \sin x \, dx.$
Evaluating the terms:

$\int_0^{\pi/2} x^2 \cos x \, dx = \frac{\pi^2}{4} - \int_0^{\pi/2} 2x \sin x \, dx.$

For the remaining integral:

$\int_0^{\pi/2} x \sin x \, dx = ( -x \cos x )_0^{\pi/2} + \int_0^{\pi/2} \cos x \, dx.$

Evaluating:

$\int_0^{\pi/2} x \sin x \, dx = 0 + 1 = 1.$

Thus:

$I = \frac{\pi^2}{4} - 2(1) = \frac{\pi^2}{4} - 2.$

Given that the integral is of the form:

$I = \left(\frac{\pi}{a}\right)^2 - \alpha,$

we compare terms and find:

$\alpha = 2.$

The Correct answer is: 2